Answer to Question #126533 in Mechanics | Relativity for Mwansa Kunda

Question #126533
Two objects are connected by a light string passing over a frictionless pulley. m1=5.00kg, m2=3.00kg, h=4.00m
The object of mass 5kg was released from rest. Using the principle of conservation of energy,
i. The speed of the 3.0kg object just as the 5.0kg object hits the ground.
ii. The maximum height to which the 3.0kg object rises.
Expert's answer

5 kg body locates 4 meters above ground. After it released it will go down because it heavier than 3 kg body. Energy conservation principle says:


(1) "m\\times a\\times h = \\frac{m\\times V^2}{2}"

where m - mass of body (5 kg), a - acceleration (not equals to gravitational acceleration in our case because we have 3kg body on other side), h - height (4 meters), V - velocity of body near ground.

acceleration could be found:

"F_{result} = F_{g5} - F_{g3}= m_{5}\\times g - m_{3} \\times g;"

"F_{result} = 5\\times 9.8 - 3\\times 9.8 = 49 - 29.4 = 19.6N";

"F_{result}=m_{5}\\times a;"

"a=\\frac{F_{result}}{m_{}}=\\frac{19.6}{5}=3.92 m\/s^2"

so, returning to formula (1) we can find V of body 5kg and body 3 kg (they a the same in this moment)

"V=\\sqrt{\\frac{m_{5}\\times a \\times h \\times 2}{m_{5}}}=\\sqrt{a \\times h \\times 2}=\\sqrt{3.92\\times 4\\times 2};"


Ansver for first question speed is 5.6 m/s

After 5 kg body hit ground, body 3kg will locate on height 8 m and continue move upward for some time, because it has some speed. To find maximum height which body 3 will have, let's use the same principle:

"m_{3}\\times g\\times h=\\frac{m_{3}\\times V^2}{2};"

"h=\\frac{V^2}{2\\times g}"

this time we use g because only force that affects body 3 kg is gravity.

"h=\\frac{5.6^2}{2\\times 9.8}=1.6m"

So maximal height of body 3 kg is 8 + 1.6 = 9.6m

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